An ideal fluid flows through a pipe made of two sections with diameters of 1.0 and 3.0 inches, respectively. What is the ratio of the speed of the fluid through the 3.0-inch section to the speed of the fluid through the 1.0-inch section?
A. 1 : 3
B. 1 : 6
C. 1 : 9
D. 3 : 1
E. 9 : 1
it is related by the square of the inverse, since the 3" pipe has a cross-section area 9 times the 1" pipe.
So, (C)
density1*area1*velocity1=density2*area2*velocity2
If the density is constant, then
area1/area2=velocity2/velocity1
(diameter1/diameter2)^2=v2/v2
To find the ratio of the speed of the fluid through the 3.0-inch section to the speed of the fluid through the 1.0-inch section, we need to use the principle of continuity. According to the principle of continuity, the flow rate of an ideal fluid should remain constant throughout a pipe of varying diameter. The flow rate is given by the equation:
A1v1 = A2v2
Where A1 and v1 are the cross-sectional area and velocity of the fluid in the 1.0-inch section, and A2 and v2 are the cross-sectional area and velocity of the fluid in the 3.0-inch section.
We can rearrange the equation to find the ratio of the speeds:
v2/v1 = A1/A2
The cross-sectional area of a pipe is proportional to the square of its diameter. So, the ratio of the speeds is inversely proportional to the square of the diameters:
v2/v1 = (D1/D2)^2
Given that the diameters are 1.0 and 3.0 inches, the ratio of the speeds is:
v2/v1 = (1.0/3.0)^2 = 1/9
Therefore, the correct answer is C. 1 : 9.
To find the ratio of the fluid speed through the two sections of the pipe, we can use the principle of continuity in fluid dynamics, which states that the volume flow rate, or the amount of fluid flowing per unit time, is constant in an incompressible fluid.
The continuity equation is given by:
A1v1 = A2v2
where A1 and A2 are the cross-sectional areas of the two sections of the pipe, and v1 and v2 are the fluid speeds through the respective sections.
From the problem statement, we have the following diameters:
d1 = 1.0 inches
d2 = 3.0 inches
The cross-sectional areas A1 and A2 can be calculated using the formula for the area of a circle:
A = πr^2
where r is the radius of the circle.
For the first section of the pipe with diameter d1 = 1.0 inches, the radius is given by:
r1 = d1/2 = 1.0/2 = 0.5 inches
Substituting this into the formula, we get:
A1 = π(0.5)^2 = π(0.25) = 0.7854 square inches
For the second section of the pipe with diameter d2 = 3.0 inches, the radius is given by:
r2 = d2/2 = 3.0/2 = 1.5 inches
Substituting this into the formula, we get:
A2 = π(1.5)^2 = π(2.25) = 7.0686 square inches
Now, we can equate the two volume flow rates using the continuity equation:
A1v1 = A2v2
Since the problem is asking for the ratio of v2 to v1, we can rearrange the equation to isolate v2:
v2/v1 = A1/A2
Substituting the values we calculated:
v2/v1 = 0.7854/7.0686
Dividing these two values, we find:
v2/v1 ≈ 0.111
Therefore, the ratio of the speed of the fluid through the 3.0-inch section to the speed of the fluid through the 1.0-inch section is approximately 1 : 9.
Thus, the correct answer is C. 1 : 9.